If the initial concentration of the reactant is doubled, the time for half

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If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is

Options

(a) zero
(b) fraction
(c) three
(d) one

Correct Answer:

zero

Explanation:

We know for a reaction of n th order. k = 1/(n-1)t [1/(a-x)ⁿ⁻¹ – 1/aⁿ⁻¹].
when t=t(1/2), a-x = a/2.
t(1/2) = 1/(n-1)K [ 1/(a/2)ⁿ⁻¹ – 1/aⁿ⁻¹] …(i).
This time for t(1/2).
When a=2a, then time period for half reaction(t) is given by,
t = 1/(n-1)k [1/(a)ⁿ⁻¹ -1/(2a)ⁿ⁻¹]..(ii).
But t = 2t(1/2) , from (i) and (ii)
t(1/2)/t = [1/(a/2)ⁿ⁻¹ – 1/aⁿ⁻¹] / (1/a)ⁿ⁻¹ – 1/(2a)ⁿ⁻¹ .
1/2 = aⁿ⁻¹ – (a/2)ⁿ⁻¹/(a/2)ⁿ⁻¹ ⨯ (2a)ⁿ⁻¹ / (2aⁿ⁻¹- aⁿ⁻¹).
1/2 = (a/2)ⁿ⁻¹ (2ⁿ⁻¹ -1) / (a/2)ⁿ⁻¹ ⨯ (2a)ⁿ⁻¹/aⁿ⁻¹ (2ⁿ⁻¹ – 1).
1/2 = (2)ⁿ⁻¹ ⇒ n = 0. Thus order of the reaction is zero.

Related Questions:

1. Which of the following gases has the lowest value of Henry’s law constant
2. Volume concentration of 1.5 N H₂O₂ is
3. The C – C bond length in ethane is
4. Bond angle in H₂O₂ is
5. In which of the following reaction, the hydrogen peroxide is a reducing agent?

Topics: Electrochemistry and Chemical Kinetics (87)
Subject: Chemistry (2512)

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